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anonymous

  • 5 years ago

solve the first order differential equation: y-x(dy/dx)=(dy/dx)y^2e^y

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  1. anonymous
    • 5 years ago
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    Rearrange the equation to put it into a standard form,\[y-(x^2+y^2e^y)\frac{dy}{dx}=0\]which is in the form, \[M(x,y)+N(x,y)\frac{dy}{dx}=0\]We check to see if it's exact. We need to ensure,\[\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x} \rightarrow 1=1\]which is true. Now,

  2. anonymous
    • 5 years ago
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    the potential function,\[\Psi(x,y)\]has that\[\frac{\partial \Psi}{\partial x}=M=y\]and\[\frac{\partial N}{\partial x}=N=-x^2-y^2e^y\]

  3. anonymous
    • 5 years ago
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    Integrate the first partial:\[\Psi(x,y)=xy+c(y)\]and then take the partial derivative of this result with respect to y,\[\frac{\partial \Psi}{\partial y}=x+c'(y)\]so that we can identify it with \[\frac{\partial \Psi}{\partial y}=-x^2-y^2e^y\]

  4. nowhereman
    • 5 years ago
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    Above you were saying \[\frac{∂N}{∂x} = 1\] while indeed it is \[-2x\]

  5. anonymous
    • 5 years ago
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    (note a mistake above dN/dx should be dPsi/dy)

  6. nowhereman
    • 5 years ago
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    I mean where you say "We check to see if it's exact."

  7. anonymous
    • 5 years ago
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    Oh - thanks nowhereman - typo with x^2...it should just be x

  8. nowhereman
    • 5 years ago
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    right :-)

  9. anonymous
    • 5 years ago
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    great - now the whole thing's polluted.

  10. anonymous
    • 5 years ago
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    \[\frac{\partial \Psi}{\partial y}=-(x+ye^y)=x+c'(y)\]

  11. anonymous
    • 5 years ago
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    so that\[c'(y)=-2x-ye^y \rightarrow c(y)=-2xy-e^y(y^2-2y+2)\]where the last integral is doable using a few integration by parts. Sub. this back in for our solution Psi to give,\[\Psi(x,y)=xy-2xy-e^y(y^2-2y+2)\]\[=-xy-e^y(y^2-2y+2)+c=K\]

  12. anonymous
    • 5 years ago
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    where c is the constant of integration, and K is a constant due to the fact that this function Psi satisfies\[\frac{d \Psi(x,y)}{dx}=\frac{\partial \Psi}{\partial x}\frac{dx}{dx}+\frac{\partial \Psi}{\partial y}\frac{dy}{dx}=\Psi_{x}+\Psi_y \frac{dy}{dx}=0\]which is the form of the equation we had to solve in the beginning.

  13. anonymous
    • 5 years ago
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    Are you seeing any mistakes, nowhereman?

  14. anonymous
    • 5 years ago
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    So \[\Psi(x,y)=-xy-e^y(y^2-2y+2)=c\]for c, some constant.

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